Two 2.0-cm-diameter disks spaced 2.0 mm apart form a parallel-plate capacitor. The electric field between the disks is 5.0×105 V/m. An electron is launched from the negative plate. It strikes the positive plate at a speed of 2.0×107 m/s. What was the electron's speed as it left the negative plate?

Ch 25: The Electric Potential

Knight Calc - Physics for Scientists and Engineers 5th Edition

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Knight Calc 5th Edition

Ch 25: The Electric Potential

문제 24b

25장, 문제 24b

Two 2.0-cm-diameter disks spaced 2.0 mm apart form a parallel-plate capacitor. The electric field between the disks is 5.0×10⁵ V/m. An electron is launched from the negative plate. It strikes the positive plate at a speed of 2.0×10⁷ m/s. What was the electron's speed as it left the negative plate?

검증된 단계별 안내

Step 1: Identify the key concepts involved in the problem. This is a parallel-plate capacitor problem involving an electron moving through an electric field. The electron gains kinetic energy as it moves from the negative plate to the positive plate due to the work done by the electric field.

Step 2: Use the work-energy principle. The work done by the electric field on the electron is equal to the change in its kinetic energy. The formula for work done is \( W = q \cdot E \cdot d \), where \( q \) is the charge of the electron, \( E \) is the electric field strength, and \( d \) is the distance between the plates.

Step 3: Write the kinetic energy equation. The change in kinetic energy is given by \( \Delta KE = KE_{final} - KE_{initial} \), where \( KE = \frac{1}{2} m v^2 \). Substitute \( KE_{final} \) and \( KE_{initial} \) into the equation.

Step 4: Relate the work done to the change in kinetic energy. Set \( W = \Delta KE \), which gives \( q \cdot E \cdot d = \frac{1}{2} m v_{final}^2 - \frac{1}{2} m v_{initial}^2 \). Rearrange this equation to solve for \( v_{initial} \).

Step 5: Substitute known values into the equation. Use \( q = -1.6 \times 10^{-19} \) C (charge of the electron), \( E = 5.0 \times 10^5 \) V/m, \( d = 2.0 \times 10^{-3} \) m, \( m = 9.11 \times 10^{-31} \) kg (mass of the electron), and \( v_{final} = 2.0 \times 10^7 \) m/s. Solve for \( v_{initial} \) algebraically.

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주요 개념

질문에 올바르게 답하기 위해 반드시 이해해야 하는 핵심 개념들은 다음과 같습니다.

Electric Field

An electric field is a region around charged particles where other charged particles experience a force. It is defined as the force per unit charge and is measured in volts per meter (V/m). In this scenario, the electric field between the capacitor plates influences the motion of the electron as it accelerates towards the positive plate.

Kinetic Energy

Kinetic energy is the energy an object possesses due to its motion, calculated using the formula KE = 0.5 * m * v^2, where m is mass and v is velocity. In the context of the electron's motion, the change in kinetic energy as it moves from the negative to the positive plate can be related to the work done by the electric field, allowing us to determine its initial speed.

Conservation of Energy

The principle of conservation of energy states that energy cannot be created or destroyed, only transformed from one form to another. In this problem, the electrical potential energy converted into kinetic energy as the electron moves through the electric field is crucial for calculating its initial speed when it leaves the negative plate.